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Real analysis limit problem.

  1. Dec 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Let a1=0, a2=1, and a(n+2)=n*a(n+1)+an/n+1

    a)Calculate the value of a6 and a7
    b)Prove that (an) converges.
    c)Show that lim an=1-e-1 when n goes to infinity.


    3. The attempt at a solution
    I got the a part and found out that a6 19/30 and a7)91/144

    part b)
    each subsequent term becomes less in value, and therefore it is converging. How can prove that it's a Cauchy sequence?

    part c)
    I'm lost here.

    Any hints are appreciated :) I'm lost on this one.
     
  2. jcsd
  3. Dec 3, 2014 #2

    RUber

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    Homework Helper

    Did you forget a parenthesis? What you wrote above does not converge.
    ## a_1=0, a_2 = 1, a_3 = 1*1+0/2 = 1, a_4 = 2*1+1/3 = 7/3, a_5 = 3*7/3+1/4 =29/4...##
    or did you mean:
    ## a_1=0, a_2 = 1, a_3 = (1*1+0)/2 = 1/2, a_4 = (2*1/2+1)/3 = 2/3, a_5 = (3*2/3+1/2)/4 =5/8, a_6 = (4*5/8 +2/3)/5=19/30?##
    Since the latter a_6 matches your post, I will assume that is correct.

    I would start by trying some of your basic convergence tests...see if you can pick out some patterns.
    Note that if ##a_{n+2}=\frac{n a_{n+1}+a_n}{n+1}##, then ##a_{n+3}=\frac{(n+1) a_{n+2}+a_{n+1}}{n+2}=\frac{(n+1) \left(\frac{n a_{n+1}+a_n}{n+1}\right)+a_{n+1}}{n+2}##.
     
  4. Dec 3, 2014 #3

    pasmith

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    Homework Helper

    This is false: you have [itex]\min(a_n,a_{n+1}) \leq a_{n+2} \leq \max(a_n,a_{n+1})[/itex] so the sequence is neither increasing nor decreasing.

    First, show that [tex]a_{n+2} - a_{n+1} = \frac{-1}{n+1}(a_{n+1} - a_n)[/tex] and hence that [tex]a_{n+1} - a_n = \frac{-(-1)^n}{n!}.[/tex] If [itex]n \geq 3[/itex] then [tex]
    |a_{n+1} - a_n| = \frac{1}{n!} < \frac{1}{2^{n-1}}.
    [/tex]
    It is straightforward to show that a sequence [itex]b_n[/itex] which satisfies [tex]
    |b_{n+1} - b_n| < \frac{1}{2^{n-1}}[/tex] for all [itex]n \geq N_0 \in \mathbb{N}[/itex] is cauchy.

    I suggested that you prove that [tex]
    a_{k+1} - a_k = -\frac{(-1)^k}{k!}.
    [/tex] You can then find an expression for [itex]a_n[/itex] from [tex]
    \sum_{k=1}^{n-1} (a_{k+1} - a_k) = -\sum_{k=1}^{n-1} \frac{(-1)^k}{k!}.
    [/tex]
     
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